A grade school is putting on its spring show. The theater seats 120 people.
Because of the demand for tickets, the school has made the following
specifications:
• The number of tickets for children will be twice as many as the number
for adults.
• Full-price adult tickets will be $24; children’s tickets will be $12.
• At least 10 of the adult tickets will be made available to seniors aged
60 and over at a 25% discount.
• Total ticket sales must be at least $1,800
How many children's tickets would be sold?
How many adult tickets would be sold (including full-priced and senior)?
How many adult tickets can be sold with the senior discount?
Let's start by assigning variables to the number of adult tickets (A) and the number of children's tickets (C).
According to the first specification, the number of children's tickets will be twice as many as the number of adult tickets: C = 2A.
The total number of tickets sold must be at least 120, so we can write: A + C ≥ 120.
Next, let's consider the ticket prices. Full-price adult tickets are $24, and children's tickets are $12. So the revenue from adult tickets is 24A, and the revenue from children's tickets is 12C.
The total ticket sales must be at least $1,800, so we can write: 24A + 12C ≥ 1800.
Now let's consider the senior discount. At least 10 adult tickets will be sold to seniors at a 25% discount. This means that (0.75)(24) = $18 will be the price for each senior ticket.
We can write an equation for the total revenue from adult tickets, including the senior discount: 18(senior tickets) + 24(non-senior tickets) = 24A.
Now let's solve the system of equations.
1) C = 2A
2) A + C ≥ 120
3) 24A + 12C ≥ 1800
4) 18(senior tickets) + 24(non-senior tickets) = 24A
From equation 1, we can substitute C = 2A into equations 2 and 3.
2) A + 2A ≥ 120
3) 24A + 12(2A) ≥ 1800
Combining like terms, we have:
3A ≥ 120
24A + 24A ≥ 1800
Simplifying:
3A ≥ 120
48A ≥ 1800
From the first inequality, we can solve for A:
A ≥ 120/3
A ≥ 40
From the second inequality, we can solve for A:
A ≥ 1800/48
A ≥ 37.5
Since A represents the number of whole adult tickets, we know it must be an integer. Therefore, A must be at least 38.
Let's consider the case where A = 38.
From equation 1: C = 2(38) = 76.
Substituting these values into equation 3 to check if the total ticket sales are at least $1800:
24(38) + 12(76) = 912 + 912 = 1824.
Since 1824 is greater than 1800, this solution is valid.
Therefore, if at least 38 adult tickets are sold, there would be 76 children's tickets sold.
The total number of tickets sold would be:
38 adult tickets + 76 children's tickets = 114 tickets.
The number of adult tickets sold with the senior discount is at least 10 (since the problem specifies "at least"), and the remaining adult tickets (38 - 10 = 28) would be sold at the full price.
Therefore, the number of adult tickets sold (including full-priced and senior) is:
10 tickets with discount + 28 full-price tickets = 38 tickets.